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Mathematical Proof of Four-Color Theorem
By Liu Ran
Abstract
The method and basic theory are far from traditional graph theory.
Maybe they are the key factor of success. K4 regions (every region is
adjacent to other 3 regions) are the max adjacent relationship, four-color
theorem is true because more than 4 regions, there must be a non-adjacent
region existing. Non-adjacent regions can be color by the same color and
decrease color consumption.
Another important three-color theorem is that the border of regions
can be colored by 3 colors. Every region has at least 2 optional colors,
which can be permuted.
1. Introduce
How many different colors are sufficient to color the regions on a
map in such a way that no two adjacent regions have the same color?
After examining a wide variety of different planar graphs, one discovers
the apparent fact that every graph, regardless of size or complexity, can
be colored with just four distinct colors.
The famous four color theorem, sometimes known as the four-color
map theorem or Guthrie's problem. There had been numerous attempts to
prove the supposition in mathematical history, but these
so-called proofs turned out to be flawed. There had been accepted proofs
that a map could be colored in using more colors than four, such as six or
five, but proving that only four colors were required was not done
successfully until 1976 by mathematicians Appel and Haken, although
some mathematicians do not accept it since parts of the proof consisted of
an analysis of discrete cases by a computer. But, at the present time, the
proof remains viable. It is possible that an even simpler, more elegant,
proof will someday be discovered, but many mathematicians think that a
shorter, more elegant and simple proof is impossible.
In the mathematical field of graph theory, a complete graph is a
simple undirected graph in which every pair of distinct vertices is
connected by a unique edge. A complete digraph is a directed graph in
which every pair of distinct vertices is connected by a pair of unique
edges (one in each direction). Kn denotes the complete graph on n
vertices. K1 through K4 are all planar graphs. However, every planar
drawing of a complete graph with five or more vertices must contain a
crossing, and the non-planar complete graph K5 plays a key role in the
characterizations of planar graphs.
2. Four color theorem
(2.1) For any subdivision of the spherical surface into non-overlapping
regions, it is always possible to mark each of the regions with one of the
numbers 1, 2, 3, 4, in such a way that no two adjacent regions receive the
same number.
In fact, if the four-color theorem is true on spherical surface, it is also
true on plane surface. Because the map is originate from sphere, and
plane surface is part of spherical surface.
3. Strategy
K4 regions (every region is adjacent to other 3 regions) are the max
adjacent relationship, four-color theorem is true because more than 4
regions, there must be a non-adjacent region existing. Non-adjacent
regions can be color by the same color and decrease color consumption.
Another important theorem is that the border of regions can be
colored by 3 colors. Every region has at least 2 optional colors, which can
be permuted.
4. Basic axiom
(4.1) Coloring the regions on a map has nothing to do with the region
shape.
This is the only one axiom in proof. It’s obviously true. Color only
depends on adjacent relationship.
Theorem (4.2)
All color solutions for boundary adjacent regions can apply to point
adjacent regions or non-adjacent regions.
We define adjacent regions as those that share a common boundary of
non-zero length. Regions, which meet at a single point or limited points,
are not considered to be "adjacent".
Because point adjacent regions are not considered to be "adjacent",
any color solution can apply to point adjacent regions, include the color
solution of boundary adjacent regions. The free degree of non-adjacent
region is limitless. So any color solution of boundary adjacent regions can
apply to point adjacent regions and non-adjacent regions.
For example:
Scenario a: non-adjacent and point adjacent
Scenario b: boundary adjacent
All color solutions for Scenario b can apply to Scenario a.
Theorem (4.3)
Any irregular regions map can transform into a circle regions map. The
color solution for circle regions map can also apply to the irregular
regions map
Because basic axiom (4.1) Þ any irregular regions map can transform
into circle-shaped, ring-shaped or fan-shaped.
If circle-shaped, ring-shaped or fan-shaped are point adjacent or
non-adjacent, transform into boundary adjacent, finally, to transform into a
circle map, ring-shaped and fan-shaped surround circle. Because of
Theorem (4.2), the color solution of map transformed can apply to the
color solution of map transforming before.
For example:
This an irregular map.
To ensure arbitrary map can be transform into circle map, first select a
circle center, second draw a line from center to region, the least region
number crossed over is the layer number of ring. From 1 to 6, the least
region number is 2, from 1 to 8, the least region number is 2, and so both
6 and 8 are in layer 3 in circle map. Other regions are in layer 2.
Layer Region
Layer 1 1
Layer 2 2,3,4,5,7
Layer 3 6,8
To ensure to preserve the adjacent relationship in transforming,
Region Adjacent region
1 2,3,4,5,7
2 1,5,6,7
3 1,4,7
4 1,3,7,8,5
5 1,2,4,6,7,8
6 2,5,7
7 1,2,3,4,5,6,8
8 4,5,7
In circle map, the necessary condition of adjacent relationship is between
2 adjacent layers, or between 2 adjacent regions (left, right) in the same
layer. Such as below:
Region 1 (layer 1) is adjacent to 2,3(layer 2). Region 3 (layer 2) is
adjacent to 2,4(layer 2).
If regions are not in the adjacent layer or more than 2 regions in the same
layer, they are sure to be not adjacent. Such as, 6(layer 3) and 1(layer 1)
are not adjacent; 2 and 4 are not adjacent in layer 2, because there are
3,4,5 in layer 2, they can’t all adjacent to 1.
With the 2 necessary condition of adjacent relationship, check the table
one item by one item.
1 2,3,4,5,7
2 1,5,6,7
3 1,4,7
4 1,3,7,8,5
5 1,2,4,6,7,8
6 2,5,7
7 1,2,3,4,5,6,8
8 4,5,7
This is the method to check region 7 and others are similar.
First, remove regions in adjacent layers.
7 2,3,4,5
Because 2,3,4,5 are in the same layer and total 4 > 2, the region 7 can’t
preserver adjacency.
To preserver adjacency, it must to cross layers by region 4, 5 or 7.
Go back to original map. Region 7 is across over region 6 to adjacent to 5,
and is across over region 3 to adjacent to 4. The final map is below.
It equal to the standard circle map below.
Transform irregular map to circle map, 1 is circle center, 6 and 8 are in
layer 2, 7 is across layer 2,3,4. Other regions are in layer 1. The boundary
adjacent relation is never changed, but some point adjacent or
non-adjacent relations are changed to boundary adjacent relation to match
the circle map transforming.
The color solution for circle map transformed can apply to irregular map
also. Region 6 has changed color, but there is no same color between
boundary adjacent regions. It is a color solution qualified.
5. Terminology
To describe conveniently, I have defined some terms in circle map.
Solution(G(n), color1,color2,…,colork) is a color solution qualified to
color all of n regions G(n) by color in {color1,color2,…,colork}.
G(n) is the n of regions in map.
Non-adjacent regions as those no point met.
For example: 1 is non-adjacent to 2 in below circle map.
Point adjacent regions as those that meet at a single point or limited
points
For example: 1 is point adjacent to 2 in below circle map.
Boundary adjacent regions as those that share a common boundary of
non-zero length.
For example: 1, 2, 3 are all boundary adjacent to other 2 regions in below
circle map.
Covered is that the least regions in upper ring have covered one nation in
lower ring. Especially, the least N regions in upper ring covering 1 nation
in lower ring calls N Covered, all the regions in upper ring covering 1
nation in lower ring calls full Covered.
For example: 1, 2, 3, 4, 5, 6 are covering 7 in below circle map, that is 6
covering.
Supported is that the least regions in lower ring have covered one nation
in upper ring. Especially, the least N regions in lower ring covering 1
nation in upper ring calls N Supported, all the regions in lower ring
covering 1 nation in upper ring calls full Supported.
For example: 1, 2, 3, 4, 5, 6 are supporting 7 in below circle map, that is 6
supporting.
Color is to color region by one or more than one colors. It is recoded as
Color(region) = { color}. If a region can be colored by more than 1 colors,
it can be recoded as Color(region) = {color1/color2…/colork}. Such as
Color(3) = {yellow/green/gray}. Region 3 is colored by {yellow} now, but
Region 3 has the freedom to color by { gree} or {gray}.
Main color is {color1} in Color(region) = {color1/color2…/colork},
which color the region in fact.
Backup color is {color2…/colork} in Color(region) =
{color1/color2…/colork}, which doesn’t color the region in fact, but
which has the freedom to color by {color2…/colork}.
Major color is to color regions in a ring with 2 alternate colors. But if
there is odd number of regions in a ring, the head and tail are the same
color.
Isolating color is to isolate the same major color with 1 another color in a
ring, which is of odd number of regions.
For example: We can see the example in below circle map.
Major color of ring 1 is white color and no isolate color, record as
Major(1) = {white}, Isolating(1) = {};
Major color of ring 2 is red and blue color and no isolate color, record as
Major(1) = {read, blue}, Isolating(1) = {};
Major color of ring 3 is black and white color and isolate color is yellow
color, record as Major(3) = {white, black}, Isolating(3) = {yellow}.
Region number is the total region number of a ring.
For example: region number of ring 3 is 7 in below circle map. Record as
Region(3) = 7.
Border regions are all the boundary regions between colored and
uncolored. It’s the frontier of regions colored.
Inner regions are the entire boundary regions closed by Border regions.
It’s the home of regions colored. For example:
Regions 1 to 11 are colored, the Border regions are marked as yellow
color, which are close border to seal all regions colored. Inner regions are
marked as gray color.
Adjacent border regions are a region being adjacent to border regions,
there are 1 or 2 regions are adjacent to the region, but not full covered.
For example:
Region 5 and 9 are adjacent to region 10, but not covered by region 10.
Region 5 and 9 are adjacent border regions.
Empty region is a point, which is not a real region, only a proving tool.
For example, 3 regions is equivalent to 3 regions and a Empty region.
Kn regions are k regions are all adjacent. Anyone of k region is adjacent
to other k-1 regions.
For example: k2 regions are in below circle map.
k3 regions are in below circle map. Anyone region is adjacent to other
two regions.
K4 regions are in below circle map. Anyone region is adjacent to other
three regions.
Graph theory has proven K4 regions are the max adjacent relationship in
planar graph.
6. Preliminary theorem
(6.1) K4 regions have only 3 scenarios in circle map.
Because in a ring, one region can only adjacent to 2 regions at most
=> there are at most 3 regions in a ring, but K4 regions have 4 regions > 3
=> K4 regions are at least in 2 rings.
If total of rings >=3, there must be one ring insulating another ring.
=>There must be 2 regions being non-adjacent.=> total of rings <=2,
Because total of rings >=2 and total of rings < 2 => total of rings =2.
Total of rings =2 and K4 regions have 4 regions => K4 regions have
only 3 scenarios in circle map. I.e.
(6.1.1) region(1) = 1, region (2) = 3.
(6.1.2) region(1) = 2, region (2) = 2.
(6.1.3) region(1) = 3, region (2) = 1.
Is there any region across rings? No. Because there is only 2 adjacent
rings, regions can keep adjacent relationship in adjacent rings, don’t’ need
to cross rings.
(6.2) K3 regions have only 3 scenarios in circle map.
Similarly, we can get 3 scenarios.
(6.2.1) region (1) = 3.
(6.2.2) region(1) = 1, region (2) = 2.
(6.2.3) region(1) = 2, region (2) = 1.
To prove conveniently, we can unify (6.1) and (6.2). For K3 regions,
we regard there is a empty region in inner regions. Then K3 regions
become K4 regions
(6.2.1) region (1) = 1 empty region , region (2) = 3.
(6.2.2) region (1) = 1+ 1 empty region, region (2) = 2.
(6.2.3) region (1) = 2+ 1 empty region, region (2) = 1.
(6.3) Three-color theorem
If color set is {1,2,3,4}, every region has at least 2 optional colors to
keep the colors of border regions being not above 3.
Such as a K4 regions, the possible color is below
When add a region, which is adjacent to 1 border region and has 3
optional colors.
When add a region, which is adjacent to 2 border regions and has 2
optional colors.
When add a region, which is adjacent to 3 border regions and has 3
optional colors.
Only explain this scenario, others are similar.
1. To be colored by {4}.
2. To be colored by {3}.
3. To be colored by {2}.
The region colored by {1} is unchanged and {2/3/4} can color the
region added.
When add a region, which is full covering all border regions and has
4 optional colors, because there is no border regions except for the region
added now.
It’s easy to verify every region in border regions has at least 2
optional colors when total of regions is not above 5.
(6.3.1) Suppose every region in border regions has at least 2 optional
colors when total of regions equals to k(k>5).
Then add the (k+1)th region, total of regions become k+1.
If the (k+1)th region is adjacent to 1 region of border regions, it has 3
optional colors.
If the (k+1)th region is adjacent to 2 colors of border regions, it has 2
optional colors.
If the (k+1)th region is adjacent to more than 2 colors of border
regions, it has at least 2 optional colors. The proof is below.
(k+1)th region’s color depends on the adjacent border regions. E.g
regions in yellow color. {1,2,3,4} – {1,2} = {3/4}
Because (k+1)th region’s color is adjacent to 3 colors, {4} is OK.
How to color {3}? First to color (k+1)th region as {3}.
Then exchange color {3} and {4}, i.e from Solution(G(k),1,2,3,4) to
Solution(G(k),1,2,4,3)
Because (6.3.1), border regions of G(k) have at least 2 optional colors.
Then border regions of G(k) being not full covered by (k+1)th region
( gray color) and colored by {4} can color to another color and this color
must in {1,2,3}. Then (k+1)th region can be colored by {3/4} and colors
of border regions of G(k+1) is still in {1,2,3}. All the impact regions are
in the inner regions of G(k).
So every region in border regions has at least 2 optional colors when
total of regions equals to k+1.
When n <4, 3 colors are sufficient.
When n = 4, 3 colors are sufficient for border regions.
When n = 5, we can verify 3 colors are sufficient for border regions.
If n = k, 3 colors are sufficient for border regions, we can deduce n =
k+1, 3 colors are sufficient for border regions.
So to all of n, in G(n) map, every region in border regions has at least
2 optional colors to keep the colors of border regions being not above 3
Some people wonder how the inner regions’ color permuted? We can
find that the optional colors depend on the adjacent border regions. For
example:
The region colored by {4/3} = {1,2,3,4} – {1,2}. When color change
from {4} to {3}, {3} need to change, it is no impact to its adjacent border
regions, but only to inner regions. For example:
Red arrow denotes no impact; green arrow denotes impact. It’s the
same as next region permuted, which is no impact on adjacent border
regions, but only inner regions.
Base on the algorithm, we can guarantee the core center is a K4
regions, which can guarantee at least 1 region with 4 optional colors, and
any impact will stop here. E.g.
All impacts are towards inner regions, can stop at the region with
more optional colors. The worst scenario is impact stop at the region with
4 optional colors, which can stop any type of impact.
More impacts are similar to 1 impact, because more impacts can be
handled one by one. The region with 4 optional colors still exists.
(6.4) K4 regions are the max adjacent relationship in circle map.
(6.4.1) Suppose K5 regions are the max adjacent relationship in circle
map.
Base on the K4 regions scenario of (6.1.1), let me to add a new
country in circle map, which is K5 regions.
(6.4.1.1) The new country is in ring 1, like below circle map.
Because they are K5 regions, country 1 is full covered by country
2,3,4 and country 5 is also full covered by country 2,3,4. 3 countries
cover both country 1 and 5. And countries 2,3,4 are in a ring, so the
overlap countries are at most 2 countries. Then the total countries in ring
2 , denoted as Country(2) >= 3+3-2 = 4>3. It’s contradictory.
(6.4.1.2) The new country is in ring 2, like below circle map.
Because countries 2,3,4,5 are in the same ring 2, country 5 can be
only Boundary adjacent with 2 countries in the same ring. But there are 3
other counties in the ring 2, so one country must be Non-adjacent with
country 5.
(6.4.1.3) The new country is in a new ring 3, like below circle map.
Country 5 is a new country in new ring. Obviously, country 5 is
always Non-adjacent with country 1, because they are in ring 1 and ring 3,
ring 2 has insulated them.
(6.4.1.4) The new country is in a new ring k, (k>3), like below circle
map.
Country 5 is a new country in new ring. Obviously, country 5 is
always Non-adjacent with country 1, because they are in ring 1 and ring k,
ring 2 has insulated them.
(6.4.1.5) The new country is across ring 2, like below circle map.
We can divide country 5 into multiple countries in each ring like
below.
Similar to scenario (6.4.1.2), because countries 2,3,4,52 are in the
same ring 2, country 52 can be only Boundary adjacent with 2 countries
in the same ring. Select one of them (such as country 4) to find another
non-adjacent country. Country 4 can be only Boundary adjacent with 2
countries in the same ring. But there are 3 other counties in the ring 2, so
one country must be Non-adjacent with country 4.
(6.4.1.6) The new country is not across ring 2, like below circle map.
Similar to scenario (6.4.1.3)
Obviously, country 5 is always Non-adjacent with country 1, because
ring 2 has insulated them.
The same method can prove scenarios of (6.1.2) and (6.1.3).
All scenarios are contradictory, so supposition (6.4.1) is false and K4
regions are the max adjacent relationship in circle map.
7. Four color theorem
There are N regions. Firstly, search max adjacent relationship.
If no K2 regions, all regions are non-adjacent, one color is sufficient.
Max adjacent relationship is K2 regions, all regions are at most
adjacent to one region, and two colors are sufficient.
Max adjacent relationship is K3 regions, we can add a empty region in
ring 1, it becomes K4 regions. border regions are formed.
Max adjacent relationship is K4 regions, because K4 regions have two
rings, border regions are formed now.
Next we select the 5th adjacent region to color. There are 3 scenarios:
Scenario 1. There is only 1 region in border regions
We can always find one non-adjacent regions in border regions,
because there are 3 regions in home of K4 regions, and there are 2 regions
in home of K3 regions . The 5th region has 3 optional colors.
Scenario 2. There are 2 regions in border regions
We can always find one non-adjacent region in border regions or
inner regions, because there are 2 regions in home of K4 regions, and
there is 1 region in home of K3 regions. The 5th region has 2 optional
colors.
Scenario 3. There are 3 regions in border regions.
Scenario 3.1. The 5th region is adjacent to less than 3 regions in border
regions.
We can always find one non-adjacent region in inner regions. The 5th
region has at least 2 optional colors.
Scenario 3.2. The 5th region is adjacent to 3 regions in border regions.
We can always find one non-adjacent region in inner regions. For K3
regions, we can find the empty regions. Because inner region can be
colored by {1,2,3,4}, The 5th region has at least 3 optional colors.
The border regions become longer and longer. When border regions are
more than 3 regions, there are 3 scenarios also. Assume to color the kth
region and border regions are above three. Colors of border regions are in
{1,2,3}.
(7.1) When the kth region is adjacent to one region in border regions. We
can always color it in {1,2,3}.
(7.2) When the kth region is adjacent to two regions in border regions. We
can always color it in {1,2,3}.
(7.3) When the kth region is adjacent to 3 colors in border regions. We can
color it by {3}.
Next to exchange color {3} and {4} in G(k).
Next to change border regions colored by {4}, which are not covered by
the kth region. Base on (6.3) three-color theorem, they can change color
from {4} to other colors in {1,2,3}, because they have at least 2 optional
colors. All the impact regions are in inner regions of G(k).
Above proof indicates border regions’ colors are in {1,2,3}, and inner
regions’ colors are in {1,2,3,4}.
Finally, four-color theorem is proven now!
8. Verification and Demo
One example to verify and explain:
To descript clearly, all regions are marked the number by the color
order in advance. The order number (region number) map is below,
The algorithm bases on (7.4), is simpler than the example in section 7.
1. Search and color max adjacent relationship of complete graph.
2. Find non-adjacent regions and colored with more possibilities.
Firstly, search K4 regions and colored by {1,2,3,4}
Next, we can select an adjacent region c5, which is adjacent to 3
regions in border. 2 adjacent border regions are {2,3}. So it is colored by
color of inner region {4/1}.
Next, we can select an adjacent region c6, which is adjacent to 2
regions in border. 2 adjacent border regions are {3,4}, so color is {2/1}.
Next, we can select an adjacent region c7, which is adjacent to 3
regions in border regions. adjacent border regions are {2,3}, so color is
{1/4}.
Next, we can select an adjacent region c8, which is adjacent to all
border regions. It is full covered all border regions. So it can be colored
by {4/1/2/3}.
Every region has at least 2 optional colors. All regions use 4 colors.
9. Conclusion
Four-color theorem is an interesting phenomenon, but there is a
rule hidden the phenomenon. The max adjacent relationship on a
surface decides how many colors are sufficient. More than max
adjacent regions, there is a non-adjacent region. The non-adjacent
region can decrease color consumption. Every region has at least 2
optional colors, which can be permuted.
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[3] LAM, T.Y. A First Course in Noncommutative Rings,
Springer-Verlag, 1991.
[4] ROTMAN, J.J. An Introduction to the Theory of Groups,
Springer-Verlag, 1995.
[5] VAN LINT, J.H. & WILSON, R.M. A Course in Combinatorics,
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